Thursday, May 24, 2018

I can't say I'm even a little bit surprised that the summit with North Korea has fallen through. I wouldn't even bother blogging about this except that back in April I expressed some cautious optimism that maybe, just maybe, Trump's bull-in-the-china-shop tactics could be working. Nothing makes me happier than having my pessimistic prophecies be proven wrong, but alas, Donald Trump seems to be every bit as catastrophically incompetent and destructive as I feared.

I would say that Trump is all hat and no cattle, except that in this case it would be an insult to hats.

[UPDATE] Have I changed my mind now that the on-again off-again summit is (apparently) on again? No, I have not. Why? Because by keeping his cool, not throwing a temper tantrum, and not offering any major concessions, Kim called Trump's bluff. And Trump, by reinstating the summit without extracting any major concessions from Kim, folded like a cheap suit. Now Kim knows that Trump desperately wants a deal. (If Obama got a Peace Prize, Trump has to have one too!) That gives Kim the upper hand in the negotiations (to put it mildly).

Trump has apparently completely lost sight of the fact that simply being in the same room with the President of the United States is a huge win for Kim. Trump, the master negotiator, is handing Kim that win with nothing at all to show for it. So no, I am not impressed.

Before analyzing this case I have to hedge: I'm pretty sure I know the right answer, but not 100%. I have actually asked one of the authors of that paper to confirm my suspicions but he's busy and so it will be a while before I expect to get an answer. If his answer turns out to be substantially different from what I say here I will definitely let you all know.

So with that disclaimer in mind, I'm quite confident that this case will turn out to have the same behavior as the case where we have a laser dimmed by a filter, and the photon emissions post-filter detected using a parametric down-converter, i.e. there will be no (first-order) interference (with one exception, which will arrive at in case 4). The reason I'm confident is that this case is structurally the same as that one: we have made a modification to the experiment that allows us to know when the photon was emitted, which allows us to determine which path the photon took by comparing departure and arrival times, and so there can be no interference. Please note that I have taken pains not to say that our (potential) knowledge of the emission time causes the interference to go away. It doesn't. What causes the loss of interference is the entanglement of the photon with something else. Entanglement is a pre-requisite for measurement, which is a pre-requisite for knowledge, so it is true that if there is (potential) knowledge that there can be no interference, but the knowledge bit is a red herring. The only reason I'm using that phraseology is that it is, sadly, ubiquitous in QM pedagogy.

So, with that out of the way, let us proceed to the interesting case: We turn the laser on and off with a duty cycle of 50% and a period that is long enough that the pulses actually do produce interference. The lengths of the arms of the interferometer are adjusted so that the Nth pulse coming from the long arm exactly coincides at the detector with the N+Kth pulse coming from the short arm for some integer K>=1.

Before I go into detail on this I want to say a word about how practical such an experiment might be. To make this work, the period of the pulses has to be long enough that the individual pulses are coherent, but short enough that we can "store" at least one them "in flight" while waiting for the next one without losing coherence. Can this actually be done? Yes, it can, at least if you believe Wikipedia. There is says that fiber lasers can be built with coherence lengths exceeding 100km. That's about 300 microseconds, which is essentially forever by the standards of quantum optics. You could actually do this experiment with a regular semiconductor laser with a coherence length of "merely" 100m. It's pretty straightforward to power-cycle a laser, produce time delays, and measure the results with events happening at nanosecond/meter scales. So this would not even push the boundaries of the state of the art.

There is no doubt what would happen in this case: you would see interference, but (and this is really important) only after the Kth cycle. Before that we know that all of the light at the detector arrived via the short arm, so there is no interference. After the Kth cycle, light is arriving from both arms, so we do get interference. There would be some transient effects at the beginning and end of each pulse, but at steady-state the interference would be easily detectable. This is predicted both by quantum theory and classical E-M theory. There is absolutely nothing interesting going on here, until, that is, you make one more little modification: in addition to a 50% duty cycle, you also dim the laser with a filter.

The outcome predicted by QM is clear: dimming the laser with a filter makes no difference. Whatever we saw when the laser was bright we should also see when the laser is dim, namely, for the first K cycles of the laser we get no interference.

How is this possible? The dramatic narrative of interference usually goes something like this: interference happens when a photon (or some other particle, it doesn't really matter) is placed in a quantum superposition, usually a quantum superposition of physical locations. The usual slogan is "the photon goes both ways". The two paths are then brought together in such a way that no which-way information is available in the final state. The result is interference.

But can this possibly be happening in this case? The two paths that the photon can take are separated by an enormous amount of time, big enough that we are able to turn the laser off and back on while we are waiting for it to travel the long way 'round through our interferometer. It's already enough of a mind bender to say that we cannot know when a particular photon was emitted when the laser is on continuously, but now we seem to be going a step further: in order to have interference, we cannot even know which power cycle a detected photon was produced by! Is it really possible for a laser to produce a photon that is in a quantum superposition across power cycles? That seems extremely improbable. Surely once you turn the laser off, the universe is committed: there's a batch of photons flying through space at the speed of light in some particular quantum mechanical configuration. Surely that configuration can't be changed by something that happens (or not -- we could choose at any time not to turn the laser back on!) in the future?

There is another possibility. Maybe the interference we see is not created by one photon interfering with itself, but rather interfering with another photon produced in a different cycle. This seems a lot more plausible, but is it actually possible? Paul Dirac, one of the founders of quantum theory, once famously wrote “...each photon then interferes only with itself. Interference between different photons never occurs.” Interestingly, while I can find this quote all over the internet, and it is invariably attributed to Dirac, I cannot locate its original source. So it's possible that this quote is apocryphal. But it doesn't matter. What matters is that this sentiment was taken seriously for decades until an experiment by Magyar and Mandel debunked it in 1963. There have been books written about, and even entitled, multi-photon interference. So it's definitely a thing.The experiment as we have described it to this point is an interesting variant on the Magyar and Mandel experiment. There they used two different lasers (actually they were masers, but it doesn't matter) to generate their photons, whereas we are using one laser and separating the production of the two photons by time using power cycles and bringing them back together using delay lines. But it amounts to the same thing. The key is that we're bring the photons back together at the same time. That's the reason that the delay time in the interferometer is an integer multiple of the cycle time on the laser, otherwise it doesn't work.So this might be a mildly interesting but not earth-shattering result. Maybe someone has even done it, I don't know.But there is one thing that should make us a little queasy about this line of thought, and that is that we cannot actually control how many photons enter the interferometer on any given cycle. We can attenuate the beam so that on average we get one photon per cycle, but that will only be an average. Some cycles we will get more than one, some cycles we will get none. If we're depending on photons to interfere with each other then we need the same number of photons each cycle, otherwise some photons won't have partners to interfere with.In fact, we can actually completely eliminate the possibility that what we see is multi-photon interference simply by making the laser even dimmer. Let's attenuate the laser to the point where, rather than of one photon every cycle, we instead get one photon every 2K cycles (or more). In other words, most of the time the interferometer is totally dark. Every now and then we will get a photon. The temporal separation between photons is now much more than the coherence time of the laser, much more than the cycle time of the laser, much more than the travel time through the long arm of the interferometer. We can make it an hour or more between photons. Theory predicts that we will still see interference! Not only that, but it will be the exact same interference pattern that we saw when the interferometer was fully illuminated.

How are we to account for that? In particular: remember how above we noted that we only saw interference after the Kth cycle? How would any given photon know whether the index of the cycle that produced it was more or less than K?

Note well that this is in no way intended to highlight a problem with QM. The outcome predicted by QM is very clear, and I would give you very long odds that this prediction is correct. The problem is only in trying to tell a story about what the fleep is going on here that involves photons being emitted by the laser. I don't see any way to do it.

In fact, I'll go one step further: AFAICT, this is a strong argument for the following remarkable conclusion (and if this holds up I think it really could be earth-shattering): the quantum wave function must be physically real because that is the only thing that could account for the K-cycle delay in the onset of interference. If anyone can see a flaw in my reasoning I would really appreciate it if you would point it out.

Tuesday, May 08, 2018

This post is (part of) the answer to a puzzle I posed here. Read that first if you haven't already.

To make this discussion concrete, let's call the time it takes for light to traverse the short (or Small) arm of the interferometer Ts, the long (or Big) arm Tb (because Tl looks too much like T1).

So there are five interesting cases here. Let's start with the easy one: we illuminate the interferometer with a laser which we turn on and leave on. In that case it's a no-brainer: the photons arrive at the business end of the interferometer first from the short path Ts seconds after turning the laser on. At this point we know (because timing) which way the photons went so there is no interference. Then at time Tb the photons arrive from the long path. All the photons are identical, so we no longer know which path they took. So interference.

Case 2 (the one the original puzzle was about): the laser is turned on and stays on, but the power is modulated (or filters are put in place) so that the light level is very low, so low that the average time of arrival between individual photons at the detector is much larger than Tb.

There are two plausible-sounding answers in this case. Plausible-sounding answer #1 is that the result is exactly the same as before: after Tb we still get interference. The equations of quantum mechanics are independent of brightness, so wherever we get interference when the light is bright, we still get interference when the light is dim.

Plausible-sounding answer #2 is that when the light is dim we can tell which way the photon went by looking at the timing. Whenever we get a detection there are two possibilities: either the photon was emitted Ts seconds earlier and took the short path, or the photon was emitted Tb seconds earlier and took the long path. So if we can tell when the photon was emitted, then there will be no interference.

But can we tell? Well (and this is the answer I was originally looking for when I composed the puzzle) it depends on exactly how we make the laser dim! There are at least two ways, and they produce different results.

The first is to put some sort of shutter in front of the laser that only lets through one photon through at a time. This is equivalent to turning the laser on only for very short periods of time. If we do it this way we will get no interference.

The other way is to do nothing to the laser itself, but rather to put a filter between the laser and the interferometer that blocks (or reflects) most of the photons. The photons are blocked by the filter at random, so there is no way to tell when a particular photon got through the filter. Hence: interference.

But suppose we tweak our setup slightly so that we can tell when a photon was transmitted by the filter. How can we do this? It is this exploration that leads to (IMHO) a profound insight.

Think about it: how do you detect a photon without destroying it? You can't! The only way to detect a photon is to have some atom absorb it, and that process destroys the photon. But there is a sneaky trick we can do: we can run the photons through a parametric down-converter (PDC). A PDC is a crystal made of some material (typically some stuff called beta-barium borate or BBO) whose atoms absorb photons at one wavelength and then re-emit them as two photons at different wavelengths. The key is that these two emissions happen at more or less the same time. So we can send one of these photons into the interferometer and use the other one to tell us when this event happened. Experimental physicists actually do experiments like this routinely. To distinguish between the two photons, the one that goes into the apparatus is called the signal photon, while the other that is measured to figure out the timing is called the idler. By measuring (and hence destroying) the idler photon we can tell when the signal photon entered the interferometer, and so we can tell which way the signal photon went (by comparing the timing of entry and exit). So we cannot have interference.

Here is the profound insight: this setup will not produce interference even if we don't actually measure the idler photon! Why? (You might want to think about that for a moment before proceeding.) Because if it did, then we could use that fact to transmit information faster than light!

Here is how we would do it: instead of measuring the idlers, we send them off (via mirrors) to some distant location (let's call it L1) At the same time, we take our interferometer and move it away from the PDC by the same distance but in the opposite direction to a location we will call L2. The distance between L1 and L2 is much greater than the length of the long arm of the interferometer.

If we could produce interference by choosing not to perform any measurements on the idlers then we could use this setup to communicate faster than light by selectively measuring the idlers or not. When we measured the idlers at L1, the interference would be destroyed at L2. And this effect would have to happen instantaneously because if it didn't then we could measure idlers at L1 and still have interference at L2, and that is impossible.

The profound conclusion is that photons emitted by a parametric down-converter do not produce interference! [1]

Those of you who have read my paper on the EPRG paradox will find this all to be familiar territory. In fact, this is the exact same conclusion that was reached in that paper, and for the exact same reason: the photons emitted by a PDC are entangled, and entangled photons do not self-interfere. The reason they don't self-interfere is that entanglement is the first step of the measurement process, and it, not measurement per se, is what destroys (first-order) interference.

This is all old news (at least 17 years old). So why is this (IMHO) cool? Because we could reach this conclusion without knowing anything about entanglement! We didn't need to invoke EPR or Bell's theorem or polarization or anything like that. All we needed was the principle that which-way information destroys interference to reach the conclusion that if there is any physical process that reliably produces multiple photons at the same time, then those photons cannot self-interfere. We have shown, without doing any math, only from elementary first principles, that entangled particles are different in some deep and profound way from non-entangled ones.

I think that's cool.

That's probably enough for one post. I'll finish up the other three cases later.

---

[1] This is not quite true. The strictly correct statement is that entangled photons do not produce first-order interference. They can and do produce second-order interference, which can only be detected by transmitting classical information from L1 to L2.

Friday, May 04, 2018

The New York Times reports that California is now world's 5th largest economy. Only the U.S. as a whole, China, Japan and Germany are bigger. On top of that the vast majority of that growth came from the coastal areas, where the liberals live.

The experiment with tax policy [in Kansas] was such a failure that a Republican controlled legislature not only voted to raise taxes, but did so over the veto of the governor.

So: in your face all you who say that high taxes and regulation kill economic growth! In fact it is, and has always been, the exact opposite. Taxes fund government and infrastructure, both of which are essential components of a robust economy.

Time to take a break from politics and sociology and geek out about quantum mechanics for a while.

Consider a riff on a Michelson-style interferometer that looks like this:

A source of laser light shines on a half-silvered mirror angled at 45 degrees (the grey rectangle). This splits the beam in two. The two beams are in actuality the same color as the original, but I've drawn them in two different colors to make the two paths easier to follow. The red beam is reflected up, the green beam passes through the mirror and continues to the right. Both beams are reflected back from whence they came by a pair of regular mirrors (the white rectangles) also mounted at 45 degree angles relative to the beams. They return to the half-silvered mirror where they are recombined and sent to a detector, which registers the presence of absence of interference. (In reality it's a tad more complicated than that, but that's a sufficiently accurate description for this thought experiment.)

What distinguishes this "riff" on a traditional Michelson interferometer is that the mirrors that reflect one of the beams (the one drawn in green) are mounted on a trolley that allows them to be moved as a unit to an arbitrary distance. We use this capability to make the distance traversed by the photons on the green path be much larger than those on the red path, large enough that there is an easily measured difference in the travel time between the two paths.

Let's call the time it takes to traverse the red path T1 and the time it takes to traverse the green path T2, with T1 much smaller than T2.

So we turn on the laser. What can we expect to see? Well, the laser beam is emitted at the speed of light (obviously). After time T1 the red-path photons arrive, but the green-path photons are still en-route, so we should see no interference. Then at time T2 the green-path photons arrive. What happens then?

From a purely electromagnetic point of view we would now expect to see interference. But do we? Here is an argument that this cannot be the case: whether we see interference or not should be independent of the brightness of the beam. So turn the brightness down to the point where the time between the emission of individual photons from the laser (and hence the arrival of photons at the detector) is much greater than T2. (Of course, the actual arrival times will be random, but we can make the average time between photons be large enough that the probability of having two photons in the interferometer at the same time is indistinguishable from zero.)

If we do this with a standard interferometer (or two-slit experiment) where the path-lengths are very nearly the same, we still see interference even when the photons go through one at a time. This is the famous and mysterious phenomenon of quantum superposition, where each individual photon "goes both ways" and interferes with itself. But with this setup, "interfering with itself" would seem to be impossible. Yes, the photon goes both ways, but how can it possibly interfere with itself when the differences in travel times between the two paths are so large? By the time the red-path-part of the photon arrives at the detector, the green-path part is still en-route to the distant mirror. Likewise, by the time the green-path-part of the photon arrives at the detector, the red-path part is long gone. So individual photons can't possibly interfere with themselves in this setup, and so large numbers of photons should not be able to produce interference either.

So there are three possibilities:

1. There is no interference after T2 (in violation of standard electromagnetic theory)

2. There is interference after T2, but it goes away if the laser is dim enough (in violation of standard QM theory), or

3. There is interference after T2 even when the laser is dim. In which case the question becomes: how?

The answer next time. It turns out that this thought experiment has some pretty profound implications with respect to the interpretation of quantum mechanics. As far as I can tell from a cursory search, I'm the first to propose it, though I would be surprised if that actually holds up to scrutiny. If anyone knows where this has been analyzed in the literature I'd appreciate a pointer.

Wednesday, May 02, 2018

Two African American men arrested at a Philadelphia Starbucks last month have reached a settlement with the city and secured its commitment to a pilot program for young entrepreneurs.

Rashon Nelson and Donte Robinson chose not to pursue a lawsuit against the city, Mike Dunn, a spokesman from the Mayor’s Office, told The Washington Post. Instead, they agreed to a symbolic payment of $1 each and asked the city to fund $200,000 for a grant program for high school students aspiring to become entrepreneurs.

Wow. I doff my hat, raise a glass, and salute you, Rashon Nelson and Donte Robinson. You took a thoroughly sucky situation and turned it into something positive. You've restored my faith in humanity.